Question

Solve for ${\cot ^{ - 1}}(3) + \operatorname{c} {\text{ose}}{{\text{c}}^{ - 1}}(\sqrt 5 ) = $
A. $\dfrac{\pi }{3}$
B. $\dfrac{\pi }{4}$
C. $\dfrac{\pi }{6}$
D. $\dfrac{\pi }{2}$

Answer 1

Hint: Assume ${\cot ^{ - 1}}(3)\,$ to be A. Find \[\cos {\text{ec }}A\] given that ${\cot ^{ - 1}}(3)\,$ is A. Write A in terms of the inverse of cosec. Convert all the inverse of cosec to inverse of sin and use the formula ${\sin ^{ - 1}}x + {\sin ^{ - 1}}y = {\sin ^{ - 1}}(x\sqrt {1 - {y^2}} + y\sqrt {1 - {x^2}} )$ to get the answer.

Formula Used:
${\text{cose}}{{\text{c}}^{ - 1}}(A) = {\sin ^{ - 1}}(\dfrac{1}{A})$,${\sin ^{ - 1}}x + {\sin ^{ - 1}}y = {\sin ^{ - 1}}(x\sqrt {1 - {y^2}} + y\sqrt {1 - {x^2}} )$, $\cos {\text{e}}{{\text{c}}^2}A - {\cot ^2}A = 1$

Complete step by step solution: 
Let ${\cot ^{ - 1}}(3)\,$ be $A$
Then, $\cot A = 3$
We know that $\cos {\text{e}}{{\text{c}}^2}A - {\cot ^2}A = 1$. Therefore,
$\cos {\text{e}}{{\text{c}}^2}A - 9 = 1$
$\cos {\text{e}}{{\text{c}}^2}A = 10$
The range of ${\cot ^{ - 1}}x$ is $(0,\pi )$. Therefore, $A$ must be in the first or second quadrant. This means that \[\cos {\text{ec }}A\] is positive.
$\cos {\text{ec }}A = \sqrt {10} $
Therefore, \[A = \cos {\text{e}}{{\text{c}}^{ - 1}}(\sqrt {10} ) = {\cot ^{ - 1}}(3)\]
We can write,
${\cot ^{ - 1}}(3) + \operatorname{c} {\text{ose}}{{\text{c}}^{ - 1}}(\sqrt 5 ) = {\text{cose}}{{\text{c}}^{ - 1}}(\sqrt {10} ) + \operatorname{c} {\text{ose}}{{\text{c}}^{ - 1}}(\sqrt 5 )$
\[ = {\sin ^{ - 1}}(\dfrac{1}{{\sqrt {10} }}) + {\sin ^{ - 1}}(\dfrac{1}{{\sqrt 5 }})\]
We know that ${\sin ^{ - 1}}x + {\sin ^{ - 1}}y = {\sin ^{ - 1}}(x\sqrt {1 - {y^2}} + y\sqrt {1 - {x^2}} )$
Therefore,
\[{\sin ^{ - 1}}(\dfrac{1}{{\sqrt {10} }}) + {\sin ^{ - 1}}(\dfrac{1}{{\sqrt 5 }}) = {\sin ^{ - 1}}(\dfrac{1}{{\sqrt {10} }}\sqrt {1 - \dfrac{1}{5}} + \dfrac{1}{{\sqrt 5 }}\sqrt {1 - \dfrac{1}{{10}}} )\]
\[ = {\sin ^{ - 1}}(\dfrac{1}{{\sqrt {10} }}\sqrt {\dfrac{4}{5}} + \dfrac{1}{{\sqrt 5 }}\sqrt {\dfrac{9}{{10}}} ) = {\sin ^{ - 1}}(\dfrac{2}{{\sqrt {50} }} + \dfrac{3}{{\sqrt {50} }})\]
$ = {\sin ^{ - 1}}(\dfrac{5}{{5\sqrt 2 }}) = {\sin ^{ - 1}}(\dfrac{1}{{\sqrt 2 }})$
We know that ${\sin ^{ - 1}}(\dfrac{1}{{\sqrt 2 }}) = \dfrac{\pi }{4}$
Therefore, the correct option is B) $\dfrac{\pi }{4}$

Note: Once we get $\cos {\text{e}}{{\text{c}}^2}A = 10$ we must be careful to not write $\cos {\text{ec }}A = \pm \sqrt {10} $ as \[\cos {\text{ec }}A\] must be positive. We can also solve this question by assuming $\cos {\text{e}}{{\text{c}}^{ - 1}}(\sqrt 5 )$ to be $A$. If we do that, we need to use the formula, ${\tan ^{ - 1}}x + {\tan ^{ - 1}}x = {\tan ^{ - 1}}(\dfrac{{x + y}}{{1 - xy}})$.